Mathematics
In the given figure, ABCD is a parallelogram, E is the mid-point of BC. DE produced meets AB produced at L. Prove that:
(i) AB = BL
(ii) AL = 2DC
Answer
(i) Given,
ABCD is a parallelogram.
We know that,
Opposite sides of a parallelogram are equal.
⇒ AB = CD and AD = BC
In △DEC and △BEL,
⇒ ∠LBE = ∠DCE (Alternate angles, since AB || DC)
⇒ EC = EB (Given)
⇒ ∠DEC = ∠BEL (Vertically opposite angles are equal)
∴ △DEC ≅ △BEL (By A.S.A axiom)
⇒ DC = BL (Corresponding parts of congruent triangles are equal)
Since, AB = DC
∴ AB = BL.
Hence, proved that AB = BL.
(ii) From figure,
⇒ AL = AB + BL
⇒ AL = DC + DC
⇒ AL = 2DC.
Hence, proved that AL = 2DC.
Related Questions
In the given figure, in △ABC, ∠B = 90°. If ABPQ and ACRS are squares, prove that:
(i) △ACQ ≅ △ABS
(ii) CQ = BS.
Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:
(i) ∠SAQ = ∠ABC
(ii) SQ = AC.
Equilateral triangle ABD and ACE are drawn on the sides AB and AC of △ABC as shown in the figure. Prove that :
(i) ∠DAC = ∠EAB
(ii) DC = BE
In the given figure, ABCD is a square and P, Q, R are points on AB, BC and CD respectively such that AP = BQ = CR and ∠PQR = 90°. Prove that:
(i) PB = QC
(ii) PQ = QR
(iii) ∠QPR = 45°
